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这是 SO 问题的后续问题：手动编写线性编程练习。我对出于教学目的实施单纯形算法（线性规划）也有类似的兴趣。我知道单纯形算法的简单实现可能有许多改进。但我有兴趣开发一个最小的完整单工代码。

到目前为止，我发现的最接近的实现是在这个答案中，由于@Pii 对上述 matlab 中的问题，为了完整起见，在下面复制。从相对较短的代码中，我可以看到涉及旋转的单纯形的基本逻辑是如何实现的（正如@RamNarasimhan 的回答中所指出的那样）。但我不知道如何处理线性规划问题的特殊情况，包括：

  1. 不可行问题，其中可行域是空的。

  2. 无界问题，其中目标函数值不受上述约束（对于标准 LP 格式：即最大化问题）

我想通过处理这两种情况，一个单纯形实现将是完整的，因为它可以处理任何 LP 问题。

我的问题是如何修改代码来处理这两种特殊情况。

注意：我不是在寻找完整的代码，只是描述处理上述两种极端情况的具体逻辑。我还查看了寻找“简单”整数线性编程源代码/伪代码，但没有找到答案。

-- @Pii 提供的代码 --

function [x, fval] = mySimplex(fun, A, B, lb, up)

%Examples paramters to show that the function actually works 

% sample set 1 (works for this data set)

% fun = [8 10 7];
% A = [1 3 2; 1 5 1];
% B = [10; 8];
% lb = [0; 0; 0];
% ub = [inf; inf; inf];

% sample set 2 (works for this data set)

fun = [7 8 10];
A = [2 3 2; 1 1 2];
B = [1000; 800];
lb = [0; 0; 0];
ub = [inf; inf; inf];


% generate a new slack variable for every row of A 

numSlackVars = size(A,1); % need a new slack variables for every row of A 

% Set up tableau to store algorithm data 
tableau = [A; -fun];

tableau = [tableau, eye(numSlackVars + 1)];

lastCol = [B;0];

tableau = [tableau, lastCol];

% for convienience sake, assign the following: 

numRows = size(tableau,1);
numCols = size(tableau,2);

% do simplex algorithm 

% step 0: find num of negative entries in bottom row of tableau 

numNeg = 0; % the number of negative entries in bottom row

for i=1:numCols 
    if(tableau(numRows,i) < 0)
        numNeg = numNeg + 1;
    end
end

% Remark: the number of negatives is exactly the number of iterations
% needed in the simplex algorithm 

for iterations = 1:numNeg 
    % step 1: find minimum value in last row 
    minVal = 10000; % some big number 
    minCol = 1; % start by assuming min value is the first element 
    for i=1:numCols
        if(tableau(numRows, i) < minVal)
            minVal = tableau(size(tableau,1), i);
            minCol = i; % update the index corresponding to the min element 
        end
    end 

    % step 2: Find corresponding ratio vector in pivot column 
    vectorRatio = zeros(numRows -1, 1);
    for i=1:(numRows-1) % the size of ratio vector is numCols - 1
        vectorRatio(i, 1) = tableau(i, numCols) ./ tableau(i, minCol);
    end 

    % step 3: Determine pivot element by finding minimum element in vector
    % ratio

    minVal = 10000; % some big number 
    minRatio = 1; % holds the element with the minimum ratio 

    for i=1:numRows-1
        if(vectorRatio(i,1) < minVal)
            minVal = vectorRatio(i,1);
            minRatio = i;
        end 
    end 

    % step 4: assign pivot element 

    pivotElement = tableau(minRatio, minCol);

    % step 5: perform pivot operation on tableau around the pivot element 

    tableau(minRatio, :) = tableau(minRatio, :) * (1/pivotElement);

    % step 6: perform pivot operation on rows (not including last row)

    for i=1:size(vectorRatio,1)+1 % do last row last 
        if(i ~= minRatio) % skip over the minRatio'th element of the tableau 
            tableau(i, :) = -tableau(i, minCol) * tableau(minRatio, :) 
                            +  tableau(i,:);
        end
    end
end 

% Now we can interpret the algo tableau 

numVars = size(A,2); % the number of cols of A is the number of variables 

x = zeros(size(size(tableau,1), 1)); % for efficiency 

% Check for basicity 
for col=1:numVars
    count_zero = 0;
    count_one = 0;
    for row = 1:size(tableau,1)
        if(tableau(row,col) < 1e-2)
            count_zero = count_zero + 1;
        elseif(tableau(row,col) - 1 < 1e-2)
            count_one = count_one + 1;
            stored_row = row; % store this column for later use 
        end
    end
    if(count_zero == (size(tableau,1) -1) && count_one == 1) % basic case
        x(col,1) = tableau(stored_row, numCols);
    else 
        x(col,1) = 0; % not basic case
    end
end

% find function optimal value at optimal solution 
fval = x(1,1) * fun(1,1); % just needed for logic to work here 
for i=2:numVars 
    fval = fval + x(i,1) * fun(1,i);
end


end
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